# Thread: Find the probability density function

1. ## Find the probability density function

A point Q is chosen in the unit square. What is the density function of the sum of the coordinates? Of the product of the coordinates?

Work-
Let X be a r.v. on (0,1) and let Y be a r.v. on (0,1). These are independent.

Then, I let V = X + Y.

So, I then perform P(V < c) where c is just some constant (cumulative distribution function). I transform the inequality like so: P(Y < c - X).

I feel like I'm very close to solving this but here is where I get stuck. What do I do now?

2. ## Re: Find the probability density function

Originally Posted by sfspitfire23
A point Q is chosen in the unit square. What is the density function of the sum of the coordinates? Of the product of the coordinates?

Work-
Let X be a r.v. on (0,1) and let Y be a r.v. on (0,1). These are independent.

Then, I let V = X + Y.

So, I then perform P(V < c) where c is just some constant (cumulative distribution function). I transform the inequality like so: P(Y < c - X).

I feel like I'm very close to solving this but here is where I get stuck. What do I do now?
Hi sfspitfire23!

The easiest way to do it, is to draw a unit square with a diagonal line representing the same values of v.
The chance P(V < c) is then the area from the origin up to this line.
This is a triangle for the first half and a diamond shaped area for the second half.

3. ## Re: Find the probability density function

In general, if X is a nonnegative r.v. with pdf f(x) and Y is a nonnegative r.v. with pdf g(x), then $P(X + Y\le c)=\int_0^c\int_0^{c-x}f(x)g(y)\,dy\,dx$. In this case, the set of points (x, y) in the unit square such that x + y <= c is a filled right triangle with vertices (0, 0), (0, c) and (c, 0), and P(X + Y <= c) is the area of that triangle. Similarly, P(XY <= c) is the area below the hyperbola y = c / x.